如果我连续多次调用小吃店,则只显示最后一个小吃栏项目。
e.g。使用下面的代码,只显示第3项。对于第1项和第2项,似乎忽略了Snackbar.LENGTH_LONG(并设置为零?)。
Snackbar.make(view, "Item 1", Snackbar.LENGTH_LONG).show();
Snackbar.make(view, "Item 2", Snackbar.LENGTH_LONG).show();
Snackbar.make(view, "Item 3", Snackbar.LENGTH_LONG).show();
但在google documents中,我发现可以对邮件进行排队。
public boolean isShownOrQueued () Returns whether this Snackbar is currently being shown, or is queued to be shown next.
那么我们如何实际排队小吃店呢?
答案 0 :(得分:6)
这是一个解决您问题的部分代码段,尽管可能没有 是正确的做事方式:
//using a queue to pass string to the snackbar
Queue<String> myQueue = new LinkedList<String>();
myQueue.offer("item 1");
myQueue.offer("item 2");
myQueue.offer("item 3");
displaysnack(myQueue, view);
public void displaysnack(final Queue dQueue, final View view){
Snackbar.make(view, (String)dQueue.poll(), Snackbar.LENGTH_LONG).setCallback(new Snackbar.Callback() {
@Override
public void onDismissed(Snackbar snackbar, int event) {
switch (event) {
case Snackbar.Callback.DISMISS_EVENT_ACTION:
Toast.makeText(getApplicationContext(), "Clicked the action", Toast.LENGTH_LONG).show();
break;
//once the timeout expires, display the next one in the queue.
case Snackbar.Callback.DISMISS_EVENT_TIMEOUT:
Toast.makeText(getApplicationContext(), "Showing: "+ (dQueue.size()), Toast.LENGTH_SHORT).show();
if (dQueue.size()>0){displaysnack(dQueue, view);}
break;
case Snackbar.Callback.DISMISS_EVENT_CONSECUTIVE:
//Toast.makeText(getApplicationContext(), "Multiple Shown", Toast.LENGTH_SHORT).show();
break;
}
}
答案 1 :(得分:4)
I also needed to implement a queue of snackbars but did not find ready solution. So I decided to implement it on my own. You can try it https://github.com/AntonyGolovin/FluentSnackbar.
Just call important()
on Builder and it will be added to the queue.
答案 2 :(得分:3)
我也实施了自己的,可能不是最适合我的需求。它在C#中用于 Xamarin 。
public class SnackbarManager : Snackbar.Callback
{
List<Snackbar> snackbarsWaiting;
List<Snackbar> snackbarsHolding;
public SnackbarManager()
{
snackbarsWaiting = new List<Snackbar>();
snackbarsHolding = new List<Snackbar>();
}
public void AddToQueue(Snackbar snackbar)
{
if (snackbar.Duration == Snackbar.LengthIndefinite) snackbar.SetDuration(Snackbar.LengthLong);
snackbar.SetCallback(this);
if (snackbarsWaiting.Count > 0 && snackbarsWaiting[0].IsShown) snackbarsHolding.Add(snackbar);
else snackbarsWaiting.Add(snackbar);
}
public void Show()
{
if (snackbarsWaiting.Count > 0 && !snackbarsWaiting[0].IsShown)
snackbarsWaiting[0].Show();
}
public override void OnDismissed(Snackbar snackbar, int evt)
{
base.OnDismissed(snackbar, evt);
snackbarsWaiting.Remove(snackbar);
if (snackbarsHolding.Count > 0)
{
snackbarsWaiting.AddRange(snackbarsHolding);
snackbarsHolding.Clear();
}
if (snackbarsWaiting.Count > 0) snackbarsWaiting[0].Show();
}
}
答案 3 :(得分:2)
我也遇到了这个问题,这是我的解决方案。
static List<Snackbar> snackBarList = new ArrayList<>();
public static void mySnackBar(CoordinatorLayout coordinatorLayout, String s,boolean queued) {
Snackbar snackbar = Snackbar.make(coordinatorLayout, s, Snackbar.LENGTH_SHORT);
if (queued) {
//if true set onDismiss CallBack
snackbar.setCallback(new Snackbar.Callback()
{
@Override
public void onDismissed(Snackbar currentSnackbar, int event) {
super.onDismissed(currentSnackbar, event);
//first remove current snackBar in List, then if List not empty show the first one
snackBarList.remove(currentSnackbar);
if (snackBarList.size() > 0)
snackBarList.get(0).show();
}
});
//add (set callback) snackBar to List
snackBarList.add(snackbar);
//the beginning
if (snackBarList.size() == 1)
snackBarList.get(0).show();
} else snackbar.show();
}
答案 4 :(得分:2)
我写了一个这样做的库。它还包括progressBar。试一试https://github.com/tingyik90/snackprogressbar